Suppose I have two random partitions of $N.$ ("random" really means "the cycle type of a random permutation", but if there is an answer with any definition, I am interested). The question is: what is the probability is that no subset sum of the first partition is equal to any subset sum of the second partition? I seem to have trouble finding references...

How are you defining subset sum? For instance $5 = 1+2+2$, do you consider all possible sums that can be generated by subsets of $\{1,2\}$ or all sums that can be generated by $1$ and two copies of $2$?
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Vidit NandaJul 12 '13 at 16:46

1 Answer
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Here are two trivial observations while we wait for the real experts to completely solve this problem (paging Prof. Stanley...)

First, note that there is a reformulation of this question that might appeal more directly to partition theorists. Let $P(n)$ be the set of all partitions of $n$. Given partitions $\pi \in P(n)$ and $\rho \in P(m)$, we write $\rho \subset \pi$ to indicate containment as multi-sets, meaning that each part of $\rho$ is also a part of $\pi$. Note that if $\rho \subset \pi$ then we must have $m \leq n$ for obvious reasons.

For each $n$, let $\Gamma(n)$ be the set of partition-pairs $(\pi_1,\pi_2) \in P(n) \times P(n)$ for which there exists a triple $(m,\rho_1,\rho_2)$ with $m < n$, $\rho_j \in P(m)$ and $\rho_j \subset \pi_j$. This question is asking:

What is $1 - \frac{|\Gamma(n)|}{|P(n)|^2}$, where $|\cdot|$ indicates cardinality?

Secondly, one can compute $|\Gamma(n)|$ by hand for small $n$. It helps a lot, at least for tiny $n$, to note that $|\Gamma(n)| \geq |P(n-1)|^2$: the latter quantity counts those partitions of $n$ which contain $1$, so no two such partitions can hope to be subset-independent. Here are some values resulting from my (possibly erroneous) calculations:

Interesting, thanks! (of course, this is a different distribution from what I had mentioned, but as I said, I will take anything :))
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Igor RivinJul 14 '13 at 19:55

@IgorRivin Sorry, I have no idea what it means to have the cycle type of a random permutation, but presumably you can weight subsets of $P(n) \times P(n)$ by a measure different from this uniform "cardinality-ratio" and proceed as before.
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Vidit NandaJul 14 '13 at 20:16